Construct a trapezium ABCD where AB | CD, AD = BC = 3.2 cm, AB = … - Vidyadip

Question

Construct a trapezium $\text{ABCD}$ where $\ce{AB \| CD, AD = BC = 3.2\ cm, AB = 6.4\ cm}$ and $\ce{CD = 9.6\ cm}$.
Measure $\angle\text{B}\ \text{and}\ \angle\text{A}.$

$[$Hint: Difference of two parallel sides gives an equilateral triangle.$]$

✓

Answer

Steps of construction:
$1.$ Draw a line segment $\text{DC} = 9.6\ cm$.
$2.$ With $D$ as centre, draw an angle measure $60^\circ$.
Now, cut$-$off it with an arc $3.2\ cm$ called point $A$.
$3.$ Now draw a parallel $\text{AB}$ to $\text{CD}$.
$4.$ Taking $C$ as centre, cut an arc $B$ measure $3.2\ cm$ on previous parallel line.
$5.$ Draw a line segment $\text{BE} = 3.2\ cm$ from arc $B$.
$6.$ Join $B$ to $E$ and $C$.
Thus, we have required trapezium $\text{ABCD}$ in which $\angle\text{A}=120^\circ$ and $\angle\text{B}=60^\circ.$

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